$ \left(\dfrac{81}{4}\right)^{-\frac{3}{2}}$
Solution: $= \left(\dfrac{4}{81}\right)^{\frac{3}{2}}$ $= \left(\left(\dfrac{4}{81}\right)^{\frac{1}{2}}\right)^{3}$ To simplify $\left(\dfrac{4}{81}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left(? \right)^{2}=\dfrac{4}{81}$ To simplify $\left(\dfrac{4}{81}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left({\dfrac{2}{9}}\right)^{2}=\dfrac{4}{81}$ so $ \left(\dfrac{4}{81}\right)^{\frac{1}{2}}=\dfrac{2}{9}$ So $\left(\dfrac{4}{81}\right)^{\frac{3}{2}}=\left(\left(\dfrac{4}{81}\right)^{\frac{1}{2}}\right)^{3}=\left(\dfrac{2}{9}\right)^{3}$ $= \left(\dfrac{2}{9}\right)\cdot\left(\dfrac{2}{9}\right)\cdot \left(\dfrac{2}{9}\right)$ $= \dfrac{4}{81}\cdot\left(\dfrac{2}{9}\right)$ $= \dfrac{8}{729}$